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\begin{document}
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\begin{center}
\vskip 1cm {\LARGE\bf A multidimensional version of a result of \\
\vskip .5cm Davenport-Erd\H{o}s}
\vskip 1cm
\large
O-Yeat Chan, Geumlan Choi, and Alexandru Zaharescu\\
Department of Mathematics\\
University of Illinois \\
1409 West Green Street\\
Urbana, IL 61801\\
USA \\
\href{mailto:ochan@math.uiuc.edu}{ochan@math.uiuc.edu}\\
\href{mailto:g-choi1@math.uiuc.edu}{g-choi1@math.uiuc.edu}\\
\href{mailto:zaharesc@math.uiuc.edu}{zaharesc@math.uiuc.edu}
\end{center}
\vskip .5in
\begin{center}
{\bf Abstract}
\end{center}
Davenport and Erd\H{o}s showed that the distribution of values
of sums of the form
\begin{equation*}
S_h(x)=\sum_{m=x+1}^{x+h} \left(\frac mp \right),
\end{equation*}
where $p$ is a prime and $\left(\frac mp \right)$ is
the Legendre symbol, is normal as $h, p\to\infty$ such
that $\frac{\log h}{\log p}\to 0$. We prove a similar
result for sums of the form
\begin{equation*}
S_h(x_1, \ldots, x_n)=\sum_{z_1=x_1+1}^{x_1+h} \cdots
\sum_{z_n=x_n+1}^{x_n+h} \left( \frac {z_1+ \cdots+z_n}{p}
\right).
\end{equation*}
\vskip .5in
\section{Introduction}\label{sec1}
Given a prime number $p$, an integer $x$ and a positive integer $h$,
we consider the sum
\[
S_h(x)=\sum_{m=x+1}^{x+h} \left(\frac mp \right),
\]
where here and in what follows $\left(\frac mp \right)$ denotes
the Legendre symbol. The expected value of such a sum is
$\sqrt h$.
If $p$ is much larger than $h$, it is a very difficult problem
to show that there is any cancellation in an individual sum
$S_h(x)$ as above. The classical inequality of
P\' olya-Vinogradov (see ~\cite{P}, ~\cite{V}) shows that
$S_h(x)=O(\sqrt p\log p)$, and assuming the Generalized
Riemann Hypothesis, Montgomery and Vaughan ~\cite{MV} proved
that $S_h(x)=O(\sqrt p\log\log p)$.
The results of Burgess ~\cite{B} provide cancellation in
$S_h(x)$ for smaller values of $h$, as small as $p^{1/4}$.
One does expect to have cancellation in $S_h(x)$ for
$h>p^{\epsilon}$, for fixed $\epsilon>0$ and $p$ large.
This would imply the well-known hypothesis of Vinogradov
that the smallest positive quadratic nonresidue mod $p$
is $

0$ and $p$ large
enough in terms of $\epsilon$. We mention that Ankeny ~\cite{A}
showed that assuming the Generalized Riemann Hypothesis,
the smallest positive quadratic nonresidue mod $p$ is
$O(\log^2p)$. It is much easier to obtain cancellation,
even square root cancellation, if one averages $S_h(x)$
over $x$. In fact, Davenport and Erd\H{o}s~\cite{DE}
entirely solved the problem of the distribution of values
of $S_h(x)$, $0\leq x

0$ such that
\begin{equation}\label{eq:3-2-2}
\left|\Phi_{n, p^{\prime}}(\lambda)-\Phi(\lambda) \right| \ge
\delta, \quad \text{for all $p^{\prime}$}.
\end{equation}
By the two theorems of Helly (see the introduction to ~\cite{S})
there exists a subsequence $\{\Phi_{n, p^{\prime \prime}} \}$ of
$\{\Phi_{n, p^{\prime}} \}$ which converges to a distribution
$\Psi$ at every point of continuity, and
\[
\int_{-\infty}^{\infty}t^r d \Psi(t)=\lim_{p^{\prime \prime} \to
\infty }\int_{-\infty}^ {\infty} t^r d \Phi_{n, p^{\prime \prime}}
=\int_{-\infty}^{\infty} t^r d \Phi (t).
\]
Since $\Phi$ is the only distribution with these special moments
$\mu_1, \mu_2, \ldots$, we have $\Psi (t)= \Phi(t)$ for all $t$.
This contradicts ~\eqref{eq:3-2-2}. Hence one concludes that,
as $p \to \infty$,
\[
\frac 1{p^n} M_{n, p}(\lambda)=
\Phi_{n, p}( \lambda) \to
\Phi(\lambda)=\frac {1}{\sqrt {2 \pi}}
\int_{-\infty}^{\lambda}e^{-\frac 12 t^2} dt,
\]
which completes the proof of the theorem.
\end{proof}
We remark that $c_n$ can be explicitly computed for any given
value of $n$. The following proposition provides an equivalent
formulation of $c_n$, which allows for easier
computations in higher dimensions. For any $n$, consider the
polynomial in two variables
\begin{equation*}
g_n(X,Y)=\sum_{l=0}^{n-1}\left(\sum_{k=0}^l (-1)^k {\binom
{n}{k}}{\binom {X+(l-k)Y+n-1}{n-1}}\right)^2.
\end{equation*}
Note that the total degree of $g_n(X,Y)$ is at most $2n-2$.
\begin{proposition}\label{pp1}
For any $n$,
\begin{equation*}
c_n = \sum_{k=0}^{2n-2} \frac{a_{n,k}}{k+1}\,,
\end{equation*}
where $a_{n,k}$ is the coefficient of $X^{k}Y^{2n-2-k}$
in $g_n(X,Y)$.
\end{proposition}
\begin{proof}
We know that for fixed $n$ and $h\to\infty$,
\begin{equation*}
\sum_{m} N_m^2 = h^{2n-1}c_n+O_n(h^{2n-2}),
\end{equation*}
where $N_m=N_m(h,n)$ is the number of
$n$-tuples $(a_1,\dots,a_n)$ such that $a_1+\cdots +a_n=m$, with
$1\leq a_i \leq h$.
Replacing $m$ by $m^\prime=m-n$ and each $a_i$ by
$b_i=a_i-1$, we get $\sum_m N_m^2 =
\sum_{m^\prime} (N^\prime_{m^\prime})^2$, where
$N^\prime_{m^\prime}$ is the number of $n$-tuples
$(b_1,\dots,b_n)$ such that $b_1+\cdots+b_n=m^\prime$,
with $0\leq b_i \leq h-1$.
Now, the number of ways to obtain a sum of $m^\prime$ from $n$
non-negative integers, with no restrictions, is $\binom
{m^\prime+n-1}{n-1}$. If we restrict any fixed $b_i$ to satisfy
the inequality $b_i\geq h$, then the number of ways drops to
$\binom {m^\prime-h+n-1}{n-1}$. If we restrict any two $b_i, b_j$
to satisfy $b_i, b_j \geq h$ then we have $\binom
{m^\prime-2h+n-1}{n-1}$ ways, and so on.
Since for each $k$, there are $\binom {n}{k}$ ways to choose
exactly $k$ of the $b_i$'s to be greater than $h$, we obtain by
the inclusion-exclusion principle,
\[
N^\prime_{m^\prime}=\sum_{0\leq k\leq m^\prime/h} (-1)^k {\binom
{n}{k}}{\binom {m^\prime-kh+n-1}{n-1}}.
\]
So we have, for $lh \leq m^\prime < (l+1)h$, $0\leq l \leq n-1$,
\[
N^\prime_{m^\prime} = \sum_{k=0}^l (-1)^k {\binom {n}{k}}{\binom
{m^\prime-kh+n-1}{n-1}}.
\]
Replacing $m^\prime$ by $s+lh$, with $0\leq s\leq h-1$,
we get
\[
N^\prime_{s+lh} = \sum_{k=0}^l (-1)^k {\binom {n}{k}}{\binom
{s+(l-k)h+n-1}{n-1}}.
\]
Therefore
\[
\sum_{m^\prime} (N^\prime_{m^\prime})^2 = \sum_{s=0}^{h-1}
\sum_{l=0}^{n-1}\bigg(\sum_{k=0}^l (-1)^k {\binom {n}{k}}{\binom
{s+(l-k)h+n-1}{n-1}}\bigg)^2
\]
\[=\sum_{s=0}^{h-1} g_n(s,h).\]
It follows that
\begin{equation}\label{AB1}
\sum_{s=0}^{h-1} g_n(s,h)=h^{2n-1}c_n+O_n(h^{2n-2}).
\end{equation}
Now, the main contribution in $g_n(s,h)$
comes from the terms where the exponents of $s$
and $h$ add up to $2n-2$. Since for any $0\leq k\leq 2n-2$,
\[
\sum_{s=0}^{h-1} s^k =
\frac{1}{k+1}h^{k+1}+O_n(h^k),
\]
we obtain
\begin{align*}
\sum_{s=0}^{h-1} g_n(s,h) &=
\sum_{s=0}^{h-1} \left(\sum_{k=0}^{2n-2} a_{n,k}s^kh^{2n-2-k} +
\text{lower order terms} \right)\\
&=\sum_{k=0}^{2n-2} \sum_{s=0}^{h-1}a_{n,k}s^kh^{2n-2-k} +
O_n(h^{2n-2})\\
&=\sum_{k=0}^{2n-2}\frac{a_{n,k}}{k+1}h^{2n-1} + O_n(h^{2n-2}).
\end{align*}
By combining this with (\ref{AB1}),
we obtain the desired result.
\end{proof}
For $n=2, 3, 4, 5, 6$, one finds that $c_2=\frac 23$,
$c_3=\frac{11}{20}$,
$c_4=\frac {151}{315}$, $c_5=\frac {15619}{36288}$,
$c_6=\frac {655177}{1663200}$. The numerator and the denominator
of $c_n$ grow rapidly as $n$ increases. For instance, for $n=10$
and $n=25$ we have
\begin{equation*}
c_{10}=\frac{37307713155613}{121645100408832}\,,
\end{equation*}
and
\begin{equation*}
c_{25}=\frac{675361967823236555923456864701225753248337661154331976453}
{34659935272607838226339154
60520201577706853740052480000000}\,.
\end{equation*}
One can also work with boxes instead of cubes, and obtain similar
distribution results. For example, in dimension two, we may
consider the sum
\[
S_{h, k}(x,y)=\sum_{u=x+1}^{x+h} \sum_{v=y+1}^{y+k} \left( \frac
{u+v}{p} \right),
\]
where $x$, $y$ are any integers and $h$, $k$ are positive integers,
with $h \ge k$, say. Then, by using the same arguments as in the
proof of Theorem~\ref{th1}, one can prove the following result.
\begin{theorem}\label{th2}
Let $h$, $k$ be functions of $p$ such that
\[
h \ge k, \quad \frac hk \to \alpha, \quad k \to \infty,
\quad \frac {\log
k}{\log p} \to 0, \quad \text{as $p \to \infty$}.
\]
Denote $\beta=\sqrt {\alpha-\frac 13}$ and $\beta^{\prime}=\sqrt
{1-\frac 1{3 \alpha}}$. Let $M_p(\lambda)$ be the number of pairs
$(x, y)$ with $0 \le x, y